Dynamic Programming for inserting R multiplication numbers

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I. Question proposal

Insert the r multiplication number (1 <= r <n <16) In a number string consisting of n numbers and divide it into R + 1 integers, find a method to insert a multiplication number to maximize the product of the R + 1 integer.

For example, for a given number of 847313926 strings, how do I insert r = five multiplication numbers to maximize the product?

 

Ii. Dynamic Planning and Design

Generally, it is not suitable to insert R multiplication numbers using exhaustive examples. Note that inserting R multiplication numbers is a multi-stage decision-making problem. It is appropriate to use dynamic planning to solve the problem.

1. Establish a recursive relationship

If f (I, k) is set, it indicates the maximum value of the product of K multiplication numbers inserted in the first I bits, a (I, j) indicates the J-I + 1 (I <= J) integer from number I to number J.

To look for Recursive Relations, let's look at an instance first: for a given 847313926, how to insert r = five multiplication numbers to maximize the product? Our goal is to obtain the optimal value F (9, 5 ).

① If four multiplication numbers have been inserted into the first eight digits, the maximum product is F (8, 4) * 6;

② If four multiplication numbers have been inserted into the first seven digits, the maximum product is F () * 26;

③ If four multiplication numbers have been inserted into the first six digits, the maximum product is F (6, 4) * 926;

④ If four multiplication numbers have been inserted into the first five digits, the maximum product is F (5, 4) * 3926;

The maximum value is F (9, 5 ).

To calculate F (8, 4 ):

① If three multiplication numbers are inserted into the first seven digits, the maximum product is F () * 2;

② If three multiplication numbers have been inserted into the first six digits, the maximum product is F (6, 3) * 92;

③ If three multiplication numbers are inserted into the first five digits, the maximum product is F (5, 3) * 392;

④ If three multiplication numbers are inserted into the first four digits, the maximum product is F (1392;

The maximum value of the above four values is F (8, 4 ).

Generally, in order to obtain f (I, k) and evaluate the first I digit of a numeric string, set the first J (k <= j <I) in the number has been inserted on the basis of the K-1 multiplication number, after the J number Insert the T multiplication number, obviously at this time the maximum product is F (J, k-1) * A (J + 1, I ). So we can get the recursive relationship:

F (I, K) = max (f (J, k-1) * a (J + 1, I) (k <= j <I)

When the first J numbers are not inserted with the multiplication number, the value is obviously an integer composed of the first J numbers. Therefore, the boundary value is:

F (J, 0) = A (1, J) (1 <= j <= I)

In the program design, the array a is omitted and replaced by the variable D.

2. Recursive calculation of the optimal value

For (D = 0, j = 1; j <= N; j ++) {<br/> d = D * 10 + B [J-1]; // assign each digit of the input numeric string to array B <br/> F [J] [0] = D; // calculate the initial value f [J] [0] <br/>}</P> <p> for (k = 1; k <= r; k ++) <br/> {<br/> for (I = k + 1; I <= N; I ++) <br/> for (j = K; j <I; j ++) {<br/> for (D = 0, u = J + 1; U <= I; U ++) <br/> d = D * 10 + B [U-1]; // calculate d as a (J + 1, I) <br/> If (F [I] [k] <F [J] [k-1] * D) // calculate f [I] [k] By recursion <br/> F [I] [k] = f [J] [k-1] * D; <br/>}< br/>}

 

3. Construct the optimal solution

In order to print the product of the inserted multiplication number, set the T (K) and C (I, K) Arrays for the location of the annotation, where C (I, K) is the position of the K th multiplication Number of the corresponding F (I, K), and T (k) indicates the position of the K th multiplication number "*", for example, T (2) = 3, indicating that the first "*" number is behind the first number.

When F (I, K) = f (J, k-1) * D is assigned to an array element, C (I, K) = J is assigned to f (I, k) The position of the k-th multiplication number is J. Based on the result that the nth multiplication position T (r) = C (n, R) = J of F (n, R) is obtained) (1 <= k <= R-1) can be applied to the next type of backward production:

T (K) = C (t (k + 1), K)

Based on the value of the T array, you can print the multiplication formula of the inserted multiplication number obtained by the surface area in the form of characters.

 

3. Program Implementation

 // Insert R multiplication numbers into a number to maximize the product <br/> // use dynamic programming to solve the problem <br/> # include <stdio. h> <br/> # include <string. h> </P> <p> int main () <br/> {<br/> char numstr [16]; <br/> int I, J, K, len, U, R, B [16], t [16], C [16] [16]; <br/> double F [17] [17], D; </P> <p> printf ("enter an integer:"); <br/> scanf ("% s", numstr ); <br/> printf ("Enter the number of inserted multiplication Numbers R:"); <br/> scanf ("% d", & R ); <br/> Len = strlen (numstr); <br/> If (LEN <= r) {<br/> printf ("the input integer is not enough or R is too large! /N "); <br/> return 0; <br/>}</P> <p> printf (" insert % d multiplication numbers in integer % S, maximum Product:/N ", numstr, R); <br/> for (D = 0, j = 0; j <= len-1; j ++) <br/> B [J] = numstr [J]-'0'; <br/> for (D = 0, j = 1; j <= Len; j ++) {<br/> d = D * 10 + B [J-1]; // convert a character in array B to a value <br/> F [J] [0] = D; // F [J] [0] assigns an initial value <br/>}</P> <p> for (k = 1; k <= r; k ++) <br/> for (I = k + 1; I <= Len; I ++) <br/> for (j = K; j <I; j ++) {<br/> for (D = 0, u = J + 1; U <= I; U ++) <br/> d = D * 10 + B [U-1]; <br/> If (F [I] [k] <F [J] [k-1] * D) {<br/> F [I] [k] = f [J] [k-1] * D; <br/> C [I] [k] = J; <br/>}</P> <p> T [R] = C [Len] [r]; <br/> for (k = r-1; k> = 1; k --) <br/> T [k] = C [T [k + 1] [k]; // reverse route the K * number. <br/> T [0] = 0; t [R + 1] = Len; <br/> for (k = 1; k <= R + 1; k ++) {<br/> for (u = T [k-1] + 1; U <= T [k]; U ++) <br/> putchar (numstr [U-1]); // output the optimal solution <br/> If (k <R + 1) <br/> putchar ('*'); <br/>}< br/> printf ("= %. 0f/N ", F [Len] [r]); // output the optimal value </P> <p> return 0; <br/>}

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// Insert R multiplication numbers into a number to maximize the product <br/> // use dynamic programming to solve the problem <br/> # include <stdio. h> <br/> # include <string. h> </P> <p> const int maxn = 41; <br/> int N [maxn], Len; </P> <p> // NT... NZ integer synthesis <br/> long NN (int t, int Z) <br/>{< br/> int I; <br/> long a = N [T]; </P> <p> for (I = t + 1; I <= z; I ++) <br/> A = A * 10 + N [I]; </P> <p> return; <br/>}</P> <p> int main (void) <br/>{< br/> int I, N, C, K, T; <br/> long num, F [maxn] [maxn]; <br/> char s [maxn]; </P> <p> scanf ("% d", & N, & C); <br/> scanf ("% s", S ); <br/> Len = strlen (s); <br/> for (I = 0; I <Len; I ++) <br/> N [I] = s [I]-'0'; <br/> K = 0; <br/> F [0] [0] = N [0]; <br/> for (I = 1; I <Len; I ++) <br/> F [I] [0] = f [I-1] [0] * 10 + N [I]; <br/> K = 1; <br/> for (k = 1; k <= C; k ++) {<br/> for (I = K; I <Len; I ++) {<br/> long a =-1; <br/> for (t = k-1; t <I; t ++) {<br/> num = nn (t + 1, I); <br/> long B = f [T] [k-1] * num; <br/> if (a <B) <br/> A = B; <br/>}< br/> F [I] [k] =; <br/>}< br/> printf ("% d/N", F [Len-1] [c]); </P> <p> return 0; <br/>}

 

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